AME 50531: Intermediate ThermodynamicsHomework Solutions
Fall 2010
1 Homework 1 Solutions
1.1 Problem 1:
CPIG air enters and isentropic nozzle at 1.30 atm and 25◦C with a velocity of 2.5 m/s. The nozzle entrance diameteris 120mm. The air exits the nozzle at 1.24atm with a velocity of 90m/s. Determine the temeprature of the exiting airand the nozzle exit diameter.
At the nozzle entrance (State 1):
P1 = 1.30atm · 101.3 kPa1 atm
= 131.7 kPa (1)
T1 = 25◦C +273.15= 298.15K (2)
v1 = 2.5 m/s (3)
A1 = π · (0.120m)2 = 0.045m2 (4)
At the nozzle exit (State 2):
P2 = 1.24atm · 01.3 kPa1 atm
= 125.6 kPa (5)
v2 = 90 m/s (6)
T2 andd2 are unknown.
Because the nozzle is isentropic, we can use the isentropic relation for temperature and pressure to find the exittemperature. For air,k = 1.4.
T2
T1=
(
P2
P1
) k−1k
(7)
T2 = T1
(
P2
P1
)k−1
k
= 298.15K ·(
125.6 kPa131.7 kPa
)1.4−1
1.4
(8)
T2 = 294.14K = 21◦C (9)
To find the exit diameter, recall that ˙m = ρAv. Then,
m = ρ1A1v1 = ρ2A2v2 (10)
Knowing thatρ1 = ρ2,A1v1 = A2v2 (11)
A2 = A1
(
v1
v2
)
= 0.045m2 ·(
2.5 m/s90 m/s
)
= 0.0012m2 (12)
Therefore, d2 = 0.020m = 20 mm .
1
1.2 Problem 2: 9.47
Consider a steam turbine power plant operation near critical pressure, as shown in Figure 1.2. As a first approximation,it may be assumed that the turbine and pump processes are reversible and adiabatic. Neglecting any changes in kineticand potential energies, calculate:
• The specific turbine work output and the turbine exit state.
• The pump work input and enthalpy at the pump exit state.
• The thermal efficiency of the cycle.
We are givenP1 = P4 = 20MPa, T1 = 800◦C, P2 = P3 = 10 kPa, andT3 = 40◦C.
State 1: Using Table B.1.3,h1 = 4069.80kJ/kg, s1 = 7.0544 kJkg·K
State 2:
s2 = s1 = 7.0544kJ
kg ·K (13)
Using Table B.1.2,s2 = 0.6492+7.5010⇒ x2 = 0.8539 (14)
h2 = 191.81+0.8539·2392.82= 2235.04kJ/kg (15)
The specific work output from the turbine,
wt = h1−h2 = 4069.80kJ/kg−2235.04kJ/kg = 1834.76kJ/kg (16)
The turbine exit state is a saturated mixture.
State 3: Compressed liquid, using Table B.1.1,h3 = 167.54kJ/kg, v3 = 0.001007
State 4: Using the property relation for constantv,
wp = −v3(P4−P3) = −0.001007· (20000−10)= −20.13kJ/kg (17)
h4 = h3−wp = 167.54kJ/kg +20.13kJ/kg = 187.67kJ/kg (18)
The heat transfer in the boiler is
qin = h1−h4 = 4069.80kJ/kg−187.67kJ/kg = 3882.13kJ/kg (19)
wnet = wt + wp = 1834.76kJ/kg−20.13kJ/kg = 1814.63kJ/kg (20)
ηT H =wnet
qin=
1814.633882.13
= 0.4674 (21)
2
2 Homework 2 Solutions:
2.1 Problem 1: 7.64
Helium has the lowest normal boiling point of any of the elements at 4.2 K. At this temperature the enthalpy ofevaporation is 83.3 kJ/kmol. A Carnot refrigeration cycle is analyzed for the production of 1kmol of liquid helium at4.2 K from saturated vapor at the same temperature. What is the work input to the refrigerator and the coefficient ofperformance for the cycle with an ambient temperature at 280K?
For the Carnot cycle the ratio of the heat transfers is the ratio of temperatures:
QL = nh f g = 1 kmol ·83.3 kJ/kmol = 83.3 kJ (22)
QH = QL ·TH
TL= 83.3 · 280
4.2= 5553.3 kJ (23)
WIN = QH −QL = 5553.33−83.3= 5470kJ (24)
β =QL
WIN=
83.35470
= 0.0152
[
=TL
TH −TL
]
(25)
3
2.2 Problem 2: 8.97
A piston/cylinder contains air at 1380K, 15 MPa, with V1 = 9 cm3, Acyl = 5 cm2. The piston is released, and justbefore the piston exits the end of the cylinder the pressure inside is 200kPa. If the cylinder is insulated, what is itslength? How much work is done by the air inside?
The cylinder is a control volume of air. It is insulated and therfore adiabatic, soQ = 0. By the continuity equation,
m2 = m1 = m (26)
From the energy equation (5.11),m(u2−u1) = Q1 2− W1 2 = − W1 2 (27)
From the entropy equation (8.37),
m(s2− s1) =
∫
dQT
+ S1 2gen = 0+ S1 2gen (28)
Pressure and temperature are known at State 1, but only pressure is known at State 2, soT2 must be obtained. Assumea reversible process.
S1 2gen = 0⇒ s2− s1 = 0 (29)
State 1: Using Table A.7,u1 = 1095.2 kJ/kg ands◦T1 = 8.5115 kJkg·K
m =P1V1
RT1=
15000·9 ·10−6
0.287·1380= 0.000341kg (30)
State 2: From the entropy equation,s2 = s1 so from Eq. 8.19,
s◦T2 = s◦T1 + R lnP2
P1= 8.5115+0.287ln
(
20015000
)
= 7.2724kJ
kg ·K (31)
Then, use Table A.7 to interpolate and findT2 = 447.2 K andu2 = 320.92kJ/kg.
V2 = V1T2P1
T1P2= 9 · 447.2 ·15000
1380·200= 218.7 cm3 (32)
⇒ L2 =V2
Acyl=
218.75
= 43.74cm (33)
w1 2 = u1−u2 = 774.3 kJ/kg (34)
W1 2 = m1w2 = 0.264kJ = 264J (35)
4
2.3 Problem 3:
CPIG air in a cylinder (V1 = 0.03 m3, P1 = 100kPa, T1 = 10·C) is compressed reversibly at constant temperature to apressure of 420kPa. Determine the entropy change, the heat transferred, and the work done. Also accurately plot thisprocess onT-S andP-V diagrams.
State 1:
T1 = 10·C = 283.15K
P1 = 100kPa
V1 = 0.03m3
State 2:
T2 = T1 = 283.15K
P2 = 420kPa
V2 =?
We know thatm1 = m2, so,
m1 = m2 =P1V1
RT=
100·0.030.287·283.15
= 0.0369kg (36)
V2 =mRT
P2=
0.0369·0.287·283.15420
= 0.00714m3 (37)
v1 =V1
m=
0.030.0369
= 0.813m3/kg (38)
v2 =V2
m=
0.007140.0369
= 0.193m3/kg (39)
S2−S1 = m
(
cp lnT2
T1−R ln
P2
P1
)
= 0.0369·(
1.004ln1−0.287ln420100
)
(40)
S2−S1 = −0.0152kJ/K (41)
Now, to find the heat transfer and work done,
Q1 2 = W1 2 = mRT lnv2
v1= 0.0369·0.287·283.15ln
0.1930.813
(42)
Q1 2 = W1 2 = −4.312kJ (43)
0.23 0.24 0.25 0.26270
280
290
300
S (kJ/K)
T (
K)
(a) T-S Diagram
0 0.01 0.02 0.03 0.040
100
200
300
400
500
P (
kPa)
V (m3)
(b) P-V Diagram
5
2.4 Problem 4: 12.97
Consider an ideal air-standard Stirling cycle with an idealregenerator. The minimum pressure and tmeperature in thecycle are 100kPa, 25◦C, the compression ratio is 11, and the maximum temperature inthe cycle is 1000◦C. Analyzeeach of the four processes in this cycle for work and heat transfer, and determine the overall perfomance of the engine.
P1 = 100kPa, T1 = T2 = 25◦C, v1v2
= 11,T3 = T4 = 1000◦C
From 1-2 at constant temperature,w1 2 = q1 2 = T1(s2− s1)
w1 2 = −RT1 lnv1
v2= −0.287·298.15ln11= −205.2 kJ/kg (44)
From 2-3 at constant volume,w2 3 = 0
q23 = CV0(T3−T2) = 0.717(1000−25)= 699kJ/kg (45)
From 3-4 at constant temperature,w3 4 = q3 4 = T1(s4− s3)
w3 4 = +RT3 lnv4
v3= 0.287·1273.15ln11= 876.2 kJ/kg (46)
From 4-1 at constant volume,w4 1 = 0
q41 = CV0(T1−T4) = 0.717(25−1000)= −699kJ/kg (47)
wNET = −205.2+0+876.2+0= 671kJ/kg (48)
Sinceq23 is supplied by−q41 (regenerator),
qH = q34 = 876.2 kJ/kg (49)
ηT H =wNET
qH=
671876.2
= 0.766 (50)
NOTE:
qH = q34 = RT3 ln11 (51)
qL = − q1 2 = RT1 ln11 (52)
ηT H =qH −qL
qH=
T3−T1
3=
9751273.15
= 0.766= Carnot efficiency (53)
6
3 Homework 3 Solutions:
3.1 Problem 1: 11:36
Consider an ideal steam reheat cycle where steam enters the high-pressure turbine at 4.0 MPa, 400◦C, amd thenexpands to 0.8 MPa. It is then reheated to 400◦C and expands to 10kPa in the low-pressure turbine. Calculate thecycle thermal efficiency and the moisture content of the steam leaving the low-pressure turbine.
State 3: High-pressure turbine entrance.
P3 = 4 MPa, T3 = 400◦C ⇒ h3 = 3213.51kJ/kg, s3 = 6.7689kJ
kg ·K (54)
State 4: High-pressure turbine exit.s4 = s3 ⇒ h4 = 2817.77kJ/kgState 5: Low-pressure turbine entrance.
P5 = 0.8 MPa, T5 = 400◦C ⇒ h5 = 3267.07kJ/kg, s5 = 7.5715kJ
kg ·K (55)
State 6: Low-pressure turbine exit. Use entropy to find the moisture contentx6:
s6 = s5 = 7.5715kJ
kg ·K ⇒ two-phase state (56)
x6 =s6− s f
s f g=
7.5715−0.64927.501
= 0.92285 (57)
h6 = h f + x6h f g = 191.81+0.92285·2392.82= 2400.02kJ/kg (58)
State 1: Pump entrance.P1 = 10 kPa, v1 = 0.00101m3/kg, h1 = 191.81kJ/kgState 2: Pump exit. Pump is reversible and adiabatic. Assume incompressible flow.
wP = v1(P2−P1) = 0.00101(4000−10)= 3.94kJ/kg (59)
h2 = h1 + wP = 191.81+3.94= 195.75kJ/kg (60)
wT,tot = h3−h4+ h5−h6 = 3213.51−2817.77+3267.07−2400.02= 1262.79kJ/kg (61)
qH1 = h3−h2 = 3213.51−195.75= 3017.76kJ/kg (62)
qH = qH1 + h5−h4 = 3017.76+3267.07−2817.77= 3467.06kJ/ jg (63)
ηCYCLE =(wT,tot −wP)
qH=
1262.79−3.943467.06
= 0.3631 (64)
7
3.2 Problem 2:
Consider a 600MW reheat-Rankine cycle steam power plant. The boiler pressure is 1000psia, the reheater pressureis 200psia, and the condenser pressure is 1psia. Both turbine inlet temperatures are 1000◦F. The water leaving thecondenser is saturated liquid. Determine the thermal efficiency of the power plant and the boiler mass flow rate inlbm/hr.
State 3: HP turbine entrance.P3 = 1000psia, T3 = 1000◦F , h3 = 1505.86Btu/lbm, s3 = 1.6530 Btulbm·R
State 4: HP turbine exit.P4 = 200 psia, s4 = s3 = 1.6530 Btulbm·R . Interpolating,h4 = 1297.58Btu/lbm
State 5: LP turbine entrance.P5 = 2000psia, T5 = 1000◦F , h5 = 1529.28Btu/lbm, s5 = 1.8425 Btulbm·R
State 6: LP turbine exit.P6 = 1 psia, s6 = s5 = 1.8425 Btulbm·R
x6 =s6− s f
s f g=
1.8425−0.13231.8461
= 0.9264 (65)
h6 = h f + x6h f g = 69.57+0.9264·1036.11= 1029.41Btu/lbm (66)
State 1: Pump entrance, saturated liquid.P1 = 1 psia, h1 = 69.57Btu/lbm, v1 = 0.016136f t3/lbm
State 2: Pump exit.P2 = 1000psia
wP = h1−h2 = −v1(P2−P1) ⇒ h2 = h1 + v1(P2−P1) (67)
Converting pressures intolb/ f t2, we getP1 = 144lb/ f t2 andP2 = 144000lb/ f t2.
wP = 0.016136(144000−144) · 1.285·10−3 Btu1 lb · f t
= 2.983Btu/lbm ⇒ h2 = 72.55Btu/lbm (68)
ηT =wNET
QIN=
(h3−h4)+ (h5−h6)+ (h1−h2)
(h3−h2)+ (h5−h−4)= 0.4235 (69)
Wcycle = 600MW · 1 Btu/s1.055·10−3 MW
= 5.687·106 Btu/s (70)
m =Wcycle
wNET=
Wcycle
(h3−h4)+ (h5−h6)+ (h−1−h2)· 3600s
1 hr= 2.9033·107 lbm/hr (71)
8
3.3 Problem 3: 11.45
A power plant with one open feedwater heater has a condenser temperature of 45◦C, a maximum pressure of 6MPa,and a boiler exit temperature of 900◦C. Extraction steam at 1MPa to the feedwater heater is mixed with the feedwaterline so the exit is saturated liquid into the second pump. Find the fraction of extraction steam flow and the two specificpump work inputs.
State 5: Boiler exit/turbine entrance.h5 = 4375.29kJ/kg, s5 = 7.8727 kJkg·K
State 6: Turbine is reversible and adiabatic:s7 = s6 = s5
P6 = 1 MPa, s6 = 7.8727kJ
kg ·K ⇒ h6 = 3569.39kJ/kg (72)
State 1: Condenser exit/pump 1 entrance.h1 = 188.42kJ/kg, v1 = 0.00101m3/kg, P1 = 9.593kPaState 2: Pump 1 exit.P2 = P6
wP1 = h2−h1 = v1(P2−P1) = 0.00101(1000−9.593)= 1.0 kJ/kg (73)
⇒ h2 = h1 + wP1 = 188.42+1.0= 189.42kJ/kg (74)
State 3: Feedwater exit.P3 = P2, h3 = 762.79,v3 = 0.001127m3/kg The extraction fraction can be expressed asy = m6
mtot. From the energy equation,(1− y)h2+ yh6 = h3
y =h3−h2
h6−h2=
762.79−189.423569.39−189.42
= 0.1696 (75)
State 4: Pump 2 exit.P4 = P5
wP2 = h4−h3 = v3(P4−P3) = 0.001127(6000−1000)= 5.635kJ/kg (76)
9
4 Homework 4 Solutions:
4.1 Problem 1: 12.22
A Brayton cycle produces 14MW with an inlet state of 17◦C, 100kPa, and a compression ratio of 17 : 1. The heatadded in the combustion is 960kJ. What are the highest temperature and mass flow rate of air, using properties fromTable A.7.1?
The specific heat varies; therefore it is necessary to go through the processes individually to find the net work andhighest temperatureT3.
State 1: T1 = 290.15K, P1 = 100kPa. From A.7.1,h1 = 290.43kJ/kg, s◦T1 = 6.83521 kJkg·K
State 2: The compression is reversible and adiabatic, so from Eq. 8.19,
s2 = s1 ⇒ s◦T2 = s◦T 1 + R lnP2
P1= 6.83521+0.287ln17= 7.64834
kJkg ·K (77)
⇒ T2 = 642.36K, h2 = 652.04kJ/kg (78)
From the energy equation with compressor work in,
wC = w1 2 = h2−h1 = 652.04−290.43= 361.61kJ/kg (79)
State 3: Energy equation for the combustor,
h3 = h2 + qH = 652.04+960= 1612.04kJ/kg (80)
⇒ T3 = 1480.33K, s◦T3 = 8.59596kJ
kg ·K (81)
State 4: The expansion is reversible and adiabatic so from Eq. 8.19,
s4 = s3 ⇒ s◦T4 = s◦T 3 + R lnP4
P3= 8.59596+0.287ln
117
= 7.78283kJ
kg ·K (82)
⇒ T4 = 728.36K, h4 = 744.14kJ/kg (83)
From the energy equation with turbine work out,
wT = h3−h4 = 1612.04−744.14= 867.9 kJ/kg (84)
The net work iswnet = wT −wC = 867.9−361.61= 506.29kJ/kg (85)
The total power requires a mass flow rate of
m =Wnet
wnet=
14000506.29
KWkJ/kg
= 27.65kg/s (86)
10
4.2 Problem 2: 12.36
A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100MW to an electric generator.The minimum temperature in the cycle is 290K, and the maximum temperature is 1600K. The minimum pressure inthe cycle is 100kPa, and the conmpressor compression ratio is 14 : 1. The compressor has an isentropic efficiency of85% and the turbine has an isentropic efficiency of 88%. Calculate the power output of the turbine. What fraction ofthe turbine output is required to drive the compressor? Whatis the thermal efficiency of the cycle?
Solve using constantCP0. For an ideal compressor,s2 = s1 ⇒ Implemented by Eq. 8.23
T2s = T1
(
P2
P1
)k−1
k
= 290(14)0.286= 616.4 K (87)
wCs = h2−h1 = CP0(T2−T1) = 1.004(616.4−290)= 327.71kJ/kg (88)
For the actual compressor,
⇒ wC =wSC
ηSC=
327.710.85
= 385.54kJ/kg = CP0(T2−T1) (89)
⇒ T2 = T1 +wC
CP0= 290+
385.541.004
= 674K (90)
For an ideal turbine,s4 = s3 ⇒ Implemented by Eq. 8.23
T4s = T3
(
P4
P3
) k−1k
= 1600
(
114
)0.286
= 752.8 K (91)
wT s = h3−h4 = CP0(T3−T4) = 1.004(1600−752.8)= 850.63kJ/kg (92)
For the actual turbine,
⇒ wT = ηST ·wST = 0.88·850.63= 748.56kJ/kg = CP0(T3−T4) (93)
T4 = T3−wt
CP0= 1600− 748.56
1.004= 854.4 K (94)
To calculate the overall net work and the cycle efficiency,
wNET = wT −wC = 748.56−385.54= 363.02kJ/kg (95)
m =WNET
wNET=
100000363.02
kWkJ/kg
= 275.47kg/s (96)
WNET = mwT = 275.47kg/s ·748.56kJ/kg = 206.203MW (97)
wC
wT=
385.54748.56
= 0.515 (98)
Energy input is from the combustor,
qH = CP0(T −3−T2) = 1.004(1600−674)= 929.7 kJ/kg (99)
ηT H =wNET
qH=
363.02929.7
= 0.3905 (100)
11
4.3 Problem 3: 12.41
An air-standard Ericsson cycle has an ideal regenerator. Heat is supplied at 950◦C and heat is rejected at 80◦C. Pressureat the beginning of the isothermal compression process is 70kPa. The heat added is 700kJ/kg. Find the compressorwork, the turbine work, and the cycle efficiency.
Identify the states:
Heat supplied at high temperature:T3 = T4 = 950◦C = 1223.15KHeat rejected at low temperature:T1 = T2 = 0◦C = 353.15KBeginning of the compression:P1 = 70kPaIdeal regenerator:
q2 3 = − q4 1 ⇒ qH = q3 4 = 700kJ/kg (101)
⇒ wT = qH = 700kJ/kg (102)
ηT H = ηCARNOT = 1− 353.151223.15
= 0.7113 (103)
wNET = ηT HqH = 0.7113·700= 497.895kJ/kg (104)
wC = qL = qH −wNET = 700−497.895= 202.105kJ/kg (105)
12
4.4 Problem 4: 12.53
An afterburner in a jet engine adds fuel after the turbine this raising the pressure and temperature due to the energy ofcombustion. Assume a standard condition of 800K, 250kPa after the turbine into the nozzle that exhausts at 95kPa.Assume the afterburner adds 425kJ/kg to that state with a rise in pressure for the same specific volume, and neglectany upstream effects on the turbine. Find the nozzle exit velocity before and after the afterburner is turned on.
Without afterburner:T1 = 800K, P1 = 200kPa, P2 = 95 kPaWith afterburner:v3 = v1, P4−95kPa
Assume reversible adiabatic nozzle flow, then constants from Eq. 8.23
T2−T1
(
P2
P1
)
(k−1)k
= 800K ·(
95250
)0.2857
= 606.8 K (106)
Energy Equation:
12
V 22 = CP(T1−T2) ⇒V2 =
√
2CP(T1−T2) =
√
2 ·1004J
kg ·K (800−606.8) K = 622.8 m/s (107)
Add theqAB at assumed constant volume then energy equation gives
T3 = T1 +qAB
Cv= 800+
4250.717
= 1392.63K (108)
v3 = v1 ⇒ P3 = P1
(
T3
T1
)
= 25kPa · 1392.63800
= 435.2 kPa (109)
Reversible adiabatic expansion, again from Eq. 8.23
T4 = T3
(
P4
P3
)
(k−1)k
= 1392.63·(
95435.2
)0.2857
= 901.5 K (110)
V2 =√
2CP(T3−T4) =
√
2 ·1004J
kg ·K (1392.63−901.5) K = 993.02m/s (111)
Comment: The real process adds some fuel that burns releasing energy so the temperature goes up and due to theconfined space then pressure goes up. As the pressure goes up the exit velocity increases altering the mass flow ratethrough the nozzle. As in most problems the real device is more complicated than we can describe with our simpleanalysis.
13
5 Homework 5 Solutions:
5.1 Problem 1: 12.70
A gasoline engine takes air in at 300K, 90 kPa and then compresses it. The combustion adds 1000kJ/kg to the airafter which the temperature is 2050K. Use variable heat capacities (Table A.7) and find the compression ratio, thecompression specific work and the highest pressure in the cycle.
Standard Otto cycle, solve using Table A.7.1.State 3: T3 = 2050K, u3 = 1725.71kJ/kgState 2: Combustion Process
u2 = u3−qH = 1725.71−1000= 725.71kJ/kg (112)
⇒ T2 = 960.5 K, s◦T 2 = 8.0889kJ
kg ·K (113)
State 1: Compression from 1 to 2:u1 = 214.36kJ/kg, s1 = 6.86926 kJkg·K , s2 = s1 ⇒ From Eq. 8.19
0 = s◦T2− s◦T1−R ln
(
P2
P1
)
= s◦T2− s◦T1−R ln
(
T2v1
T1v2
)
(114)
= 8.0089−6.86926−0.287ln
(
960.5300
)
−0.287ln
(
v1
v2
)
(115)
Solving⇒ v1
v2= 21.89 (116)
Comment: This is much too high for an actual Otto cycle.
− w1 2 = u2−u1 = 725.71−214.36= 511.35kJ/kg (117)
Highest pressure is after combustion wherev3 = v2 so we get
P3 = P2T3
T2= P1
(
T3
T1
)(
v1
v3
)
= 90·(
2050300
)
·21.89= 13.46MPa (118)
14
5.2 Problem 2: 12.90
A diesel engine has a state before compression of 95kPa, 300 K, a peak pressure of 6000kPa, and a maximumtemperature of 2400K. Find the volumetric compression ratio and the thermal efficency. Use the properties fromTable A.7.
Compression:State 1: T1 = 300K, u1 = 214.36,s1 = 6.869 kJ
kg·KState 2: s2 = s1 ⇒ From Eq. 8.19
s◦T 2 = s◦T1 + R ln
(
P2
P1
)
= 6.869+0.287ln
(
600095
)
= 8.0588kJ
kg ·K (119)
A.7.1⇒ T2 = 935.32K,h2 = 972.90kJ/kg (120)
State 3: h3 = 2755.78kJ/kg, s ◦T3 = 9.19586 kJ
kg·K
qH = h3−h2 = 2755.78−972.90= 1782.88kJ/kg (121)
CR =v1
v2=
(
T1
T2
)(
P2
P1
)
=
(
300935.32
)(
600095
)
= 20.26 (122)
Expansion process
s◦T4 = s◦T 3 + R ln
(
P4
P3
)
= s◦T3 + R ln
(
T4
T3
)
+ R ln
(
v3
v4
)
(123)
v3
v4=
v3
v1=
(
v2
v1
)
·(
T3
T2
)
=
(
T3
T2
)(
1CR
)
=
(
2400935.32
)(
120.26
)
= 0.1267 (124)
s◦T 4−R ln
(
v3
v4
)
= s◦T3 + R ln
(
T4
T3
)
= 9.1958+0.287ln0.1267= 8.6029 (125)
Trial and error onT4 since it appears both ins◦T4 and the ln function
T4 = 1300K, LHS = 8.4405−0.287ln
(
13002400
)
= 8.6160 (126)
T4 = 1250K, LHS = 8.3940−0.287ln
(
12502400
)
= 8.5812 (127)
Now linearly interpolate⇒ T4 = 1280.7 K, u4 = 1004.81kJ/kg
qL = u4−u1 = 1004.81−214.36= 790.45kJ/kg (128)
η = 1−(
qL
qH
)
= 1−(
790.451782.88
)
= 0.5566 (129)
15
5.3 Problem 3: 11.122
Consider an ideal dual-loop heat-powered refrigeration cycle using R-134a as the working fluid, as shown in Figure5.3. Saturated vapor at 90◦C leaves the boiler and expands in the turbine to the ocndenserpressure. Saturated vapor at−15◦C leaves the evaporator and is compressed to the condenser pressure. The ratio of the flows through the two loopsis such that the turbine produces juse enough power to drive the compressor. The two exiting streams mix togetherand enter the condenser. Saturated liquid leaving the compressor at 45◦C is then separated into two streams in thenecessary proportions. Determine the ratio of mass flow ratethrough the power loop the the refrigeration loop. Findalso the performance of the cycle, in terms of the ratioQL
QH.
From Table B.5.1,T6 = 90◦C sat. vapor⇒ P5 = P6 = 3.2445MPaFrom Table B.5.1,T3 = 45◦C sat. liquid⇒ P2 = P3 = P7 = 1.1602MPa
T1 = −15◦C, h1 = 389.20kJ/kg, h3 = h4 = 264.11kJ/kg, h6 = 425.70kJ/kg (130)
C.V. Turbine
s7 = s6 = 1.6671kJ
kg ·K = 1.2145+ x7 ·0.4962, x7 = 0.912 (131)
h7 = 264.11+0.912·157.85= 408.09kJ/kg (132)
C.V. Compressors2 = s1 = 0.9258⇒ h2 = 429.89kJ/kg (133)
CV: turbine + compressorContinuity Eq.:m1 = m2,m6 = m7 (134)
Energy Eq.: ˙m1h1 + m6h6 = m2h2 + m7h7 (135)
m6
m1=
(h2−h1)
(h6−h7)=
429.89−389.20425.20−408.09
= 2.31 (136)
CV: pumpwP = v3(P5−P3) = 0.000890(3244.5−1160.2)= 1.855kJ/kg (137)
h5 = h3 + wP = 265.97kJ/kg (138)
CV: evaporator⇒ QL = m1(h1−h4)CV: boiler⇒ QL = m6(h6−h5)
β =QL
QH=
m1(h1−h4)
m6(h6−h5)=
389.20−264.112.31(425.70−265.97)
= 0.339 (139)
16
5.4 Problem 4: 12.112
A small utility gasoline engine of 250cc runs at 1500RPM with a compression ratio of 7 : 1. The inlet state is 75kPa,17◦C and the combustion adds 1500kJ/kg to the charge. This engine runs a heat pump using R-410a with ahighpressure of 4MPa and an evaporator operating at 0◦C. Find the rate of heating the pump can deliver. Overall cycle
efficiency is from Eq. 12.12,rv = v1v2
= 7
ηT H = 1− r1−k = 1−7−0.4 = 0.5408 (140)
wNET = ηT H ·qH = 0.5408·1500= 811.27kJ/kg (141)
We also need specific volume to evaluate Eqs. 12.9 to 12.11
v1 =RT1
P1=
0.287 kJkg·K ·290K
75 kPa= 1.1097m3/kg (142)
v2 =v1
CR= 0.15853m3/kg (143)
Pme f f =wNET
v1− v2=
811.271.097−0.15853
kK/kgm3/kg
= 852.9 kPa (144)
Now we can find the power from Eq. 12.11 (assume 4-stroke engine)
W = Pme f f VdisplRPM60
12
= 852.9 ·2.5 ·10−4 · 150060
· 12
= 2.665kW (145)
For the refrigeration cycle we have:State 1: h1 = 279.12kJ/kg, s1 = 1.0368 kJ
kg·KState 2: P2 = 4 MPa, s2 = s1, interpolate⇒ h2 = 323.81kJ/kgState 3: P3 = 4 MPa, interpolate⇒ h3 = 171,62kJ/kg
βHP =qH
wC=
h2−h3
h2−h1=
323.81−171.62323.81−279.12
= 3.405 (146)
The work out of the heat engine queals the input to the heat pump
QH = β HPW = 3.405·2.665kW = 9.07kW (147)
17
6 Homework 6 Solutions:
6.1 Problem 1: 13.20
A 100 m3 storage tank with fuel gases is at 300K, 100kPa containing a mixture of acetyleneC2H2, propaneC3H8
and butaneC4H10. A test shows the partial pressure ofC2H2 is 15kPa and that ofC3H8 is 65kPa. How much mass isthere of each component?
Assume ideal gases, then the ratio of partial to total pressure is the mole fraction,y = PPtot
yC2H2 =15100
= 0.15, yC3H8 =65100
= 0.65, yC4H10 =20100
= 0.20 (148)
ntot =PVRT
=100kPa ·100m3
8.31451 kJkmol·K ·300K
= 4.0091kmol (149)
mC2H2 = (nM)C2H2 = yC2H2ntotMC2H2 = 0.15·4.0091kmol ·26.038kg/kmol = 15.659kg (150)
mC3H8 = (nM)C3H8 = yC3H8ntotMC3H8 = 0.65·4.0091kmol ·44.097kg/kmol = 114.905kg (151)
mC4H10 = (nM)C4H10 = yC4H10ntotMC4H10 = 0.20·4.0091kmol ·58.124kg/kmol = 46.604kg (152)
18
6.2 Problem 2: 13.29
A flow of 1 kg/s argon at 300K and another flow of 1kg/s CO2 at 1600K both at 200kPa are mixed without anyheat transfer. Find the exitT ,P using variable specific heats.
No work implies no pressure change for a simple flow.
Pe = 200kPa (153)
The energy equation becomes
mhi = mhe = (mhi)Ar +(mhi)CO2 = (mhe)Ar +(mhe)CO2 (154)
⇒ mCO2(he −hi)CO2 + mArCPAr(Te −Ti)Ar = 0 (155)
⇒ 1 kg/s · (he−1748.12kJ/kg)+ (1 ·0.52) kW/K · (Te −300K) = 0 (156)
heCO2+0.52Te = 1748.12+0.52·300= 1904.12kJ/kg (157)
Trial and error onTe using Table A.8 forheCO2
Te = 1200K : LHS = 1223.34+0.52·1200= 1847.34 (158)
Te = 1300K : LHS = 1352.28+0.52·1300= 2028.28 (159)
Interpolating,
Te = 1200+1001904.12−1847.342028.28−1847.34
= 1231.4 K (160)
19
6.3 Problem 3: 13.46
A mixture of 2kg oxygen and 2kg argon is in an insulated piston cylinder arrangement at 150kPa, 300K. The pistonnow compresses the mixture to half its initial volume. Find the final pressure, temperature, and the piston work.
C.V. Mixture. Control mass, boundary work, adiabatic and assume reversible.
Energy Eq. 5.11:u2−u1 = q1 2− w1 2 = − w1 2 (161)
Entropy Eq. 8.37:s2− s1 = 0+0= 0 (162)
Process: cosntants ⇒ Pvk = constant,v2 = v12 ,
Assume ideal gases(T1 À Tc) and usekmix andCvmix for properties.
Eq. 13.15:Rmix = ∑ciRi = 0.5 ·0.25983+0.5·0.20813= 0.234kJ
kg ·K (163)
Eq. 13.23:CPmix = ∑ciCPi = 0.5 ·0.9216+0.5·0.5203= 0.721kJ
kg ·K (164)
Cvmix = CPmix −Rmix = 0.487kJ
kg ·K (165)
Ratio of specific heats:kmix = CPmixCvmix
= 1.4805 The relations for the polytropic process
Eq. 8.25:P2 = P1
(
v1
v2
)k
= P1(2)k = 150(2)1.4805= 418.5 kPa (166)
Eq. 8.24:T2 = T1
(
v1
v2
)k−1
= T1(2)k−1 = 300(2)0.4805= 418.6 K (167)
Work from the energy equation
W1 2 = mtot(u1−u2) = mtotCv(T1−T2) = 4 kg ·0.487kJ
kg ·K (300−418.6) K = −231kJ (168)
20
6.4 Problem 4: 13.54
Carbon dioxide gas at 320K is mixed with nitrogen at 280K in an insulated mixing chamber. Both flows are at100kPa and the mass ratio of carbon dioxide ot nitrogen is 2 : 1. Find the exit temprature and total entropy generationperkg of the exit mixture.
CV mixing chamber. The inlet ratio is so ˙mCO2 = 2mN2 and assume no external heat transfer, no work involved.
Continuity Eq. 6.9: ˙mN2 +2mN2 = mex = 3mN2 (169)
Energy Eq. 6.10: ˙mN2(hN2 −2hCO2) = 3mN2hmixex (170)
Take 300K as reference and writeh = h300+CPmix(Tmixex −300)
CPN2(TiN2−300)+2CPCO2(TiCO2
−300) = 3CPmix(Tmixex −300) (171)
CPmix = ∑ciCPi =23·0.842+
13·1.042= 0.9087
kJkg ·K (172)
3CPmixTmixex = CPN2TiN2
+2CPCO2TiCO2
= 830.64kJ/kg (173)
Tmixex = 304.7 K (174)
To find the entropies we need the partial pressures, which assuming ideal gas are equal to the mole fractions timesthe total pressure:
yi =
[
ciMi
]
∑ c jM j
(175)
yN2 =
[ 0.333328.013
]
0.333328.013+ 0.6666
44.01
= 0.44 (176)
yCO2 = 1− yN2 = 0.56 (177)
Sgen = mexsex − (ms)iCO2 − (ms)iN2 = mN2(se − si)N2 +2mN2(se − si)CO2 (178)
Sgen
3mN2
=13
[
CPN2ln
Tex
TiN2
−RN2 lnyN2
]
+23
[
CPCO2ln
Tex
TiCO2
−RCO2 lnyCO2
]
(179)
=13
[
1.042ln
(
304.7280
)
−0.2968ln0.44
]
+23
[
0.842ln
(
304.7320
)
−0.1889ln0.56
]
(180)
Sgen = 0.1561kJ
kgmix K(181)
21
7 Homework 7 Solutions:
7.1 Problem 1: 13.73
Ambient moist air enters a steady-flow air-conditioning unit at 105kPa, 30◦C, with a 60% relative humidity. Thevolume flow rate entering the unit is 100L/s. The moist air leaves the unit at 95kPa, 15◦C, with a relative humidityof 100%. Liquid condensate also leaves the unit at 15◦C. Determine the rate of heat transfer for this process.
State 1:PV1 = φ1PG1 = 0.60·4.246kPa = 2.5476kPa (182)
w1 = 0.622· 2.5476105−2.5476
= 0.0154668 (183)
mA =PA1V1
RAT1=
102.45·0.10.287·303.15
= 0.1178kg/s (184)
State 2:PV2 = PG2 = 1.705kPa (185)
w2 = 0.622· 1.70595−1.705
= 0.1137 (186)
Energy Eq. 6.10:QCV + mAhA1 + mV1hV1 = mAhA2 + mV2hV2 + m3hL3 (187)
QCV
mA= CP0A(T2−T1)+ w2hV2−w1hV1 +(w1−w2)hL3 (188)
= 1.004(15−30)+0.1137·2528.91−0.0154668·2556.25+0.00456·63.0= −26.732kJ/kg air (189)
QCV = 0.1178(−26.732)= −3.088kW (190)
22
7.2 Problem 2: 13.103
Use a psychrometric chart to find the missing property of:φ , ω , Twet , Tdry
a. Tdry = 25◦C, φ = 80%b. Tdry = 15◦C, φ = 100%
c. Tdry = 20◦C, ω = 0.008d. Tdry = 25◦C, Twet = 23◦C
a. 25◦C, φ = 80%⇒ ω = 0.016, Twet = 22.3◦C (191)
b. 15◦C, φ = 100%⇒ ω = 0.0106, Twet = 15◦C (192)
c. 20◦C, ω = 0.008⇒ φ = 57%, Twet = 14.4◦C (193)
d. 25◦C, Twet = 23◦C ⇒ ω = 0.017, φ = 86% (194)
23
7.3 Problem 3: 14.35
The Joule-Thomson coefficientµJ is a measure of the direction and magnitude of the temperature change with pressurein a throttling process. For any three propertiesx,y,z use the mathematical relation
(
∂x∂y
)
z
(
∂y∂ z
)
x
(
∂ z∂x
)
y= −1 (195)
to show the following relations for the Joule-Thomson coefficient:
µJ =
(
∂T∂P
)
h=
T(
∂v∂T
)
P− v
CP=
RT 2
PCp
(
∂Z∂T
)
P(196)
Let x = T , y = P andz = h and substitute into the relations as(
∂T∂P
)
h
(
∂P∂h
)
T
(
∂h∂T
)
P= −1 (197)
Then we have the definition of specific heat asCP =(
∂h∂T
)
Pso solve for the first term
µJ =
(
∂T∂P
)
h= −
1CP
(
∂P∂h
)
T
= − 1CP
(
∂h∂P
)
T(198)
The last derivative is substituted with Eq. 14.25 so we get
µJ =
(
∂T∂P
)
h=
T(
∂v∂T
)
P− v
CP(199)
If we use the compressibility factor then we get
Pv = ZRT ⇒(
∂v∂T
)
P=
ZRP
+RTP
(
∂Z∂T
)
P=
vT
+RTP
(
∂Z∂T
)
P(200)
so then
T
(
∂v∂T
)
P− v = v +
RT 2
P
(
∂Z∂T
)
P− v =
RT 2
P
(
∂Z∂T
)
P(201)
and we have shown the last expression also.
µJ =
(
∂T∂P
)
h=
T(
∂v∂T
)
P− v
CP=
RT 2
PCp
(
∂Z∂T
)
P(202)
24
7.4 Problem 4: 14.77
Developgeneral expressions for changes in internal energy, enthalpy, and entropy for a gas obeying the Redlich-Kwong equation of state.
Redlich-Kwong equation of state:P =RT
v−b− a
v(v + b)T1/2(203)
(
∂P∂T
)
v=
Rv−b
+a
2v(v + b)T3/2(204)
From Eq. 14.30
u2−u1 =
∫ 2
1Cv(T,v)dT +
∫ 2
1
3a
2v(v + b)T1/2=
∫ 2
1Cv(T ,v)dT − 3a
2bT 1/2ln
[(
v2 + bv2
)(
v1
v1 + b
)]
(205)
We find the change inh from change inu, so we do not do the derivative in Eq. 14.27. This is due to the form of theEOS.
h2−h1 =
∫ 2
1CP(T ,v)dT + P2v2−P1v1−
3a
2bT 1/2ln
[(
v2 + bv2
)(
v1
v1 + b
)]
(206)
Entropy follows from Eq. 14.35
s2− s1 =
∫ 2
1Cv(T,v)
dTT
+
∫ 2
1
[
Rv−b
+a/2
v(v + b)T3/2
]
dv (207)
s2− s1 =
∫ 2
1Cv(T ,v)
dT
T+ R ln
(
v2−bv1−b
)
− a
2bT3/2ln
[(
v2 + bv2
)(
v1
v1 + b
)]
(208)
25
7.5 Problem 5: 14.81
A flow of oxygen at 235K, 5 MPa is throttled to 100kPa in a steady flow process. Find the exit temperature andthe specific entropy generation using the Redlich-Kwong equation of state and ideal gas heat capacity. Notice thisbecomes iterative due to the non-linearity couplingh, P¡ v andT .
C.V. Throttle. Steady single flow, no heat transfer and no work.
Energy Eq.:h1 +0 = h2 +2 so constanth (209)
Entropy Eq.:s1 + sgen = s2 so entropy generation (210)
Find the change inh from Eq. 14.26 assumingCP is constant.
Redlich-Kwong equation of state:P =RT
v−b− a
v(v + b)T1/2(211)
(
∂P∂T
)
v=
Rv−b
+a
2v(v + b)T3/2(212)
From Eq. 14.31
(u2−u1)T =
∫ 2
1
3a
2v(v + b)T1/2=
−3a
2bT 1/2ln
[(
v2 + bv2
)(
v1
v1 + b
)]
(213)
We find change inh from change inu, so we do not do the derivative in Eq. 14.27. This is due to the form of the EOS.
(h2−h1)T = P2v2−P1v1−3a
2bT 1/2ln
[(
v2 + bv2
)(
v1
v1 + b
)]
(214)
Entropy follows from Eq. 14.35
(s2− s1)T =
∫ 2
1
[
Rv−b
+a/2
v(v + b)T3/2
]
dv = R ln
(
v2−bv1−b
)
− a
2bT 3/2ln
[(
v2 + bv2
)(
v1
v1 + b
)]
(215)
Pc = 5040kPa, Tc = 154.6 K, R = 0.2598kJ
kg ·K (216)
b = 0.08664RTc
Pc= 0.08664· 0.2598·154.6
5040= 0.0006905m3/kg (217)
a = 0.42748R2T 5/2
c
Pc= 0.42748· (0.2598)2 · (154.6)5/2
5040= 1.7013 (218)
We need to findT2 so the energy equation is satisfied
h2−h1 = h2−hx + hx −h1 = Cp(T2−T1)+ (h2−h1)T = 0 (219)
and we will evaluate it similar to figure 13.4, where the first term is done from state x to 2 and the second term is donefrom state 1 to state x (atT1 = 235K). We do this as we assume state 2 is close to ideal gas, but we donot knowT2.
We first need to findv1 from the EOS, so guessv and findP
v1 = 0.011m3/kg ⇒ P = 5059 too high (220)
v1 = 0.01114m3/kg ⇒ P = 5000.6 OK (221)
Now evaluate the change inh along the 235K from state 1 to state x, that requires a value forvx. Guess ideal gas atTx = 235K,
vx =RTx
P2= 0.2598· 235
100= 0.6105m3/kg (222)
26
From the EOS:P2 = 99.821kPa (close) A few more guesses and adjustments gives
vx = 0.6094m3/kg, P2 = 100.0006kPa OK (223)
(hx −h1)T = Pxvx −P1v1−3a
2bT 1/2ln
[(
vx + bvx
)(
v1
v1 + b
)]
= 19.466kJ/kg (224)
From energy eq.:T2 = T1 =(hx −h1)T
Cp= 235− 19.466
0.9055= 213.51K (225)
Now the change ins is done in a similar fashion,
sgen = s2− s1 = (sx − s1)T + s2− sx = R ln
(
vx −bv1−b
)
− a
2bT 3/2ln
[(
vx + bvx
)(
v1
v1 + b
)]
+Cp lnT2
Tx(226)
= 1.0576+0.0202−0.0868= 0.9910kJ
kg ·K (227)
27
7.6 Problem 6:
Consider a thermodynamic system in which there aretwo reversible work modes: compression and electrical. So takethe version of the text’s Eq. 4.16 givingdW to be
δW = PdV −E dZ, (228)
whereE is the electrical potential difference anddZ is the amount of charge that flows into the system.
• Extend the Gibbs equation to account for electrical work.
• Find the Legendre transformation which renders the independent variables to beP, E andT and show how theother variables can be determined as functions of these independent variables.
• Find all Maxwell equations associated with this Legendre transformation.
Gibbs equation:dU = δQ−δW, δQ = Tds (229)
dU = TdS−PdV +E dZ (230)
Legendre transformation:
ψ1 = T, ψ2 = −P, ψ3 = E , x1 = S, x2 = V, x3 = Z (231)
F1 = U −ψ1x1 = U −TS (232)
F1 = U −ψ2x2 = U −PV (233)
F1 = U −ψ3x3 = U −E Z (234)
F1,2,3 = U −TS + PV −E Z (235)
F(P,E ,T ) = −SdT +VdP−ZdE (236)
Maxwell equations:
−S =
(
∂U∂T
)
P,E, −V =
(
∂U∂P
)
T,E
, −Z =
(
∂U∂E
)
T,P(237)
(
∂S∂P
)
T,E
= −(
∂V∂T
)
P,E(238)
(
∂S∂PE
)
T,P=
(
∂Z∂T
)
P,E(239)
(
∂V∂E
)
T,P= −
(
∂Z∂P
)
T,E
(240)
28
7.7 Problem 7: 14.113
A 2 kg mixture of 50% argon and 50% nitrogen by mass is in a tank at 2MPa, 180K. How large is the volume usinga model of (a) ideal gas and (b) Redlich-Kwong equation of state with a, b for a mixture.
a) Ideal gas mixture:
Eq. 13.15:Rmix = ∑ciRi = 0.5 ·0.2081+0.5·0.2968= 0.25245kJ
kg ·K (241)
V =mRmixT
P=
2 ·0.25245·1802000
= 0.0454m3 (242)
b) Redlich-Kwong equation of state:Before we can do the parameters a, b for the mixture we need theindividual component parameters, Eq. 14.54, 13.55.
aAr = 0.42748R2T 5/2
C
Pc= 0.42748
(0.2081)2 · (150.8)2.5
4870= 1.06154 (243)
aN2 = 0.42748R2T 5/2
C
Pc= 0.42748
(0.2081)2· (126.2)2.5
3390= 1.98743 (244)
bAr = 0.08664RTC
Pc= 0.08664
0.2081·150.84870
= 0.000558 (245)
bN2 = 0.08664RTC
Pc= 0.08664
0.2081·126.23390
= 0.000957 (246)
Now the mixture parameters are from Eq. 14.84
amix =(
∑cia1/2i
)2=
(
0.5 ·√
1.06154+0.5 ·√
1.98743)2
= 1.4885 (247)
bmix = ∑cibi = 0.5 ·0.000558+0.5·0.000957= 0.000758 (248)
Using now Eq. 14.53:
P =RT
v−b− a
v(v + b)T1/2(249)
2000=0.25245·180v = 0.000758
− 1.4885
v(v +0.000758)1801/2(250)
By trial and error we find the specific volume,v = 0.02102m3/kg
V = mv = 0.04204m3 (251)
29
8 Homewrok 8 Solutions:
8.1 Problem 1: 15.20
Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion of ethanol,C2H5OH.
Reaction Eq.:C2H5OH + νO2(O2 +3.76N2) ⇒ aCO2 + bH2O+ cN2 (252)
Do the atom balanceBalanceC: 2 = a (253)
BalanceH: 6 = 2b ⇒ b = 3 (254)
BalanceO: 1+2νO22a + b = 4+3= 7→ νO2 = 3 (255)
(
AF
)
mol=
νO2(1+3.76)1
= 3 ·4.76= 14.28 (256)
(
AF
)
mass=
νO2MO2 + νN2MN2
MFuel=
3 ·31.999+11.28·28.01346.069
= 8.943 (257)
30
8.2 Problem 2: 16.20
A container has liquid water at 15◦C, 100kPa in equilibrium with a mixture of water vapor and dry air also at 15◦C,100kPa. How much is the vater vapor pressure and what is the saturated water vapor pressure?
From the steam tables we have for a saturated liquid:
Pg = 1.705kPa , v f = 0.001001m3/kg (258)
The liquid is at 100kPa so it is compressed liquid still at 15◦Cso from Eq. 14.15 at constantT
gliq −g f =
∫
vdP = v f (P−Pg) (259)
The vapor in the moist air is at the partial pressurePv also at 20◦C so we assume ideal gas for the vapor
gvap −gg =
∫
vdP = RT lnPv
Pg(260)
We have two saturated phases sog f −gg (q = h f g = T s f g) and now for equilibrium the two Gibbs functions mustbe the same as
gvap = gliq = RT lnPv
P+ g+ gg = v f (P−Pg)+ g f (261)
leaving us with
lnPv
Pg=
v f (P−Pg)
RT=
0.001001(100−1.705)0.4615·288.15
= 0.0007466 (262)
Pv = Pge0.0007466= 1.7063kPa (263)
This is only a minute amount above the saturation pressure. For moist air applications such differences wereneglected and assumed the partial water vapor pressure at equilibrium (100% relative humidity) isPg. The pressurehas to be much higher for this to be asignificant difference.
31
8.3 Problem 3: 16.24:
Calculate the equilibrium constant for the reactionH2 ⇒ 2H at a temperature of 2000K, using properties from TableA.9. Compare the result with the value listed in Table A.11.
From Table A.9 at 2000K we find:
∆hH2 = 52942kJ/kmol, sH2 = 188.419kJ
kmol ·K , h◦f = 0 (264)
∆hH2 = 35375kJ/kmol, sH2 = 154.279kJ
kmol ·K , h◦f = 217999 (265)
∆G◦ = ∆H −T∆S = HRHS −HLHS −T(S◦RHS −S◦LHS) (266)
= 2 · (35375−217999)−52943−2000(2·154.279−182.419)= 213528kJ/kmol (267)
lnK =−∆G◦
RT=
−2135288.3145·2000
= −12.8407 (268)
K = 2.6507·10−6 (269)
Table A.11: lnK = −12.841 OK
32
8.4 Problem 4: 16.34
Pure oxygen is heated from 25◦C to 3200K in a steady flow process at a constant pressure of 300kPa. Find the exitcomposition and the heat transfer.
The only reaction will be the dissociation of oxygen
O2 ⇔ 2O, From A.11:K(3200) = e−3.069= 0.046467 (270)
Look at initially 1 mol Oxygen and shift reaction withx
nO2 = 1− x, nO = 2x, ntot = 1+ x, yi =ni
ntot(271)
K =y2
O
y2O2
(
PP0
)2−1
=4x2
(1+ x)2
1+ x1− x
·3 =12x2
1− x2 (272)
x2 =K/12
1+ K/12⇒ x = 0.06211 (273)
yO2 =1− x1+ x
= 0.883 , yO = 1− yO2 = 0.117 (274)
q = nO2exhO2ex + nOexhOex − hO2in = (1+ x)(yO2hO2 + yOhO)−0 (275)
hO2 = 106022kJ/kmol, hO = 249170+60767= 309937kJ/kmol ⇒ q = 137947kJ/kmol O2 (276)
q =q32
= 4311kJ/kg (277)
33
9 Homework 9 Solutions:
9.1 Problem 1: 15:70
An isobaric combustion process receives gaseous benzeneC6H6 and air in a stoichiometric ratio atP0, T0. To limit theproduct temperature to 2200k, liquid water is sprayed in after the combustion. Find thekmol of liquid water adder perkmol of fuel and the dew point of the combined products.
The reaction for stoichiometric mixture withC andH balance done is:
C6H6 + νO2(O2 +3.76N2) → 3H2O+6CO2+ cN2 (278)
O balance: 2νO2 = 3+6 ·2= 15⇒ vO2 = 7.5 (279)
N balance:c = 3.76νO2 = 3.76·7.5= 28.2 (280)
With x kmol of water added perkmol fuel the products are
Products:(3+ x)H2O+6CO2+28.2N2 (281)
Energy Eq.:HR = H◦R + xh f
◦H2O liq = H◦
P + ∆HP = H◦P + xh f
◦H2O vap +(3+ x)∆hH2O +6hCO2 +28.2hN2 (282)
Where the extra water is shown explicitly. Rearrange to get
H◦R −H◦
P −6∆hCO2 −28.2∆hN2 −3∆hH2O = x(h f◦H2O vap − h f
◦H2O liq + ∆hH2O) (283)
40756·78.114−6 ·103562−28.2·63362−3·83153= x[−241826− (−285830)+83153] (284)
525975= x(127157)⇒ x = 4.1364kmol/kmol f uel (285)
Dew point:yV =3+ x
6+28.2+ x= 0.1862 (286)
⇒ PV = yV P = 0.1862·101.325= 18.86kPa (287)
B.1.2 : Tdew = 58.7◦C (288)
34
9.2 Problem 2: 15.81
A stoichimetric mixture of benzeneC6H6 and air is mixed from the reactants flowing at 25◦C, 100 kPa. Find theadiabatic flame temperature. What is the error if constant specific heat atT0 for the products from Table A.5 are used?
C6H6 + νO2O2 +3.76νO2N2 → 3H2O+6CO2+3.76νO2N2 (289)
νO2 = 6+32
= 7.5⇒ νN2 = 28.2 (290)
HP = H◦P + ∆HP = HR = H◦
R ⇒ (291)
∆HP = −H◦RP = 40576·78.114= 3169554kJ/kmol (292)
∆HP = 6hCO2 +3hH2O +28.2hN2 (293)
∆HP 2600K = 6(128074)+3(104520)+28.2(77963)= 3280600 (294)
∆HP 2400K = 6(115779)+3(93741)+28.2(70640)= 2968000 (295)
Linear interpolation⇒ TAD = 2529K
∑νiCPi = 6 ·0.842·44.01+3·1.872·18.015+28.2·1.042·28.013= 1146.66kJ
kmol ·K (296)
∆T =∆HP
∑νiCPi=
31695541146.66
= 2764 (297)
⇒ TAD = 3062K, 21% high (298)
35
9.3 Problem 3: 16.49
Water from the combustion of hydrogen and pure oxygen is at 3800 K and 50kPa. Assuming we only haveH2O, O2
andH2 as gases, find the equilibrium composition.
With only the given components we have the reaction
2H2O ⇔ 2H2 + O2 (299)
which at 3800K has an equilibrium constant from A.11 of lnK = −1.906.Assume we start with 2kmol water and let it dissociatex to the left then
Species H2O H2 O2
Initial 2 0 0Change −2x 2x xFinal 2−2x 2x x
Total: 2+ x
Then we have
K = e−1.906=y2
H2yO2
y2H2O
(
PP0
)2+1−2
=
( 2x2+x
)2 x2+x
(2−2x2+x
)2
50100
(300)
which reduces to
0.148674=1
(1− x)2
4x3
2+ x14
12
or x3 = 0.297348(1− x)2(2+ x) (301)
Trial and error to solve forx = 0.54, then the concentrations are
yH2O =2−2x2+ x
= 0.362, yO2 =x
2+ x= 0.213, yH2 =
2x2+ x
= 0.425 (302)
36
9.4 Problem 4: 16.54
A tank contains 0.1 kmol hydrogen and 0.1 kmol of argon gas at 25◦C, 200kPa and the tank keeps constant volume.To whatT should it be heated to have a mole fraction of atomic hydrogen, H, of 8%?
For the reactionH2 ⇔ 2H, K =y2
H
yH2
(
PP0
)2−1
(303)
Assume the dissociation shifts right with an amountx, then we getAssume we start with 2kmol water and let it dissociatex to the left then
Reaction H2 ⇔ 2H also, ArInitial 0.1 0 0.1
Change −x 2x 0Equilibrium 0.1− x 2x 0.1
Total: 0.2+ x
yH =2x
0.2+ x= 0.08⇒ x = 0.008333 (304)
We need to findT soK will take on the proper value; sinceK depends onP we need to evaluateP first.
P1V = n1RT1, P2V = n2RT2 ⇒ P2 = P1n2T2
n1T1(305)
where we haven1 = 0.2 andn2 = 0.2+ x = 0.2083
K =y2
H
yH2
(
PP0
)2−1
=(2x)2
(0.1− x)n2
200100
n2T2
0.2 ·298.15= 0.0001016T2 (306)
Now it is trial and error to getT2 so the above equation is satisfied withK from A.11 atT2.
3400K : lnK = −1.519, K = 0.2189, RHS = 0.34544, error = −0.1266 (307)
3600K : lnK = −0.611, K = 0.5428, RHS = 0.36576, error = 0.1769 (308)
Linear interpolation between the two to make zero error
T = 3400+200· 0.17690.1769+0.1266
= 3483.43K (309)
37
9.5 Problem 5:
Consider the reaction of heptane and air:
ν ′1C7H16+ ν ′
2(O2 +3.76N2) ⇀↽ ν ′′3CO2 + ν ′′
4 H2O+ ν ′′5CO+ ν ′′
6 NO+ ν ′′7 NO2 + ν ′′
8 N2 (310)
Find the most general set of stoichiometric coefficients forthe reaction.
CHON
7 0 1 0 1 0 0 016 0 0 2 0 0 0 00 2 2 1 1 1 2 00 7.52 0 0 0 1 1 2
ν1
ν2
ν3
ν4
ν5
ν6
ν7
ν8
= /0 (311)
Row echelon form:
1 0 0 0 1/22 −34/1019 −16/203 14/5790 1 0 0 0 25/188 25/188 25/940 0 1 0 15/22 103/441 1141/2068 −175/10340 0 0 1 −4/11 67/251 326/517 −53/274
{
ν}
= /0 (312)
1 0 0 00 1 0 00 0 1 00 0 0 1
ν1
ν2
ν3
ν4
=
− 122ν5
341019ν6
16203ν7 − 14
579ν8
0 − 25188ν6 − 25
188ν7 − 2594ν8
− 1522ν5 − 103
441ν6 − 11412068ν7
1751034ν8
411ν5 − 67
251ν6 − 326517ν7
53274ν8
(313)
From there the four stoichiometric equations can be writtenout
ν1 = − 122
ν5 +34
1019ν6 +
16203
ν7−14579
ν8 (314)
ν2 = − 25188
ν6−25188
ν7−2594
ν8 (315)
ν3 = −1522
ν5−103441
ν6−11412068
ν7 +1751034
ν8 (316)
ν4 =411
ν5−67251
ν6−326517
ν7 +53274
ν8 (317)
38
10 Homework 10 Solutions:
10.1 Problem 1:
Plot uR versusT ∈ [300K,5000K] for He, Ar, H, O, H2, O2, H2O, andH2O2. Use thechemkin software package to
generate values of ¯u(T ).
0 1000 2000 3000 4000 50000
2000
4000
6000
8000
10000
12000
Temperature [K]
u He/R
[K]
Helium
0 1000 2000 3000 4000 50000
2000
4000
6000
8000
10000
12000
Temperature [K]
u Ar/R
[K]
Argon
0 1000 2000 3000 4000 50002.6
2.8
3
3.2
3.4
3.6
3.8x 10
4
Temperature [K]
u H/R
[K]
Hydrogen, Monatomic
0 1000 2000 3000 4000 50002.8
3
3.2
3.4
3.6
3.8
4
4.2x 10
4
Temperature [K]
u O/R
[K]
Oxygen, Monatomic
39
0 1000 2000 3000 4000 50000
0.5
1
1.5
2
2.5x 10
4
Temperature [K]
u H2/R
[K]
Hydrogen, Diatomic
0 1000 2000 3000 4000 50000
0.5
1
1.5
2
2.5x 10
4
Temperature [K]
u O2/R
[K]
Oxygen, Diatomic
0 1000 2000 3000 4000 5000−3
−2.5
−2
−1.5
−1
−0.5
0
0.5x 10
4
Temperature [K]
u H2O
/R [K
]
Water
0 1000 2000 3000 4000 5000−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
3x 10
4
Temperature [K]
u H2O
2/R [K
]
Hydrogen Peroxide
40
10.2 Problem 2:
Consider a Hydrogen dissociation reaction,H2 + H2 ⇀↽ 2H + H2 (318)
For this reaction
a = 2.23·1012 cm3 ·K− 12
mole · s (319)
β = 0.5 (320)
E = 92600cal
mole(321)
Consider an isochoric, isothermal reaction in whichnH2 = 1 kmole, nH = 0 kmole at t = 0, and for whichT =5200K andP = 1500kPa.
a) Formulate the reaction kinetics in the formdρH2
dt= f (ρH2) (322)
b) Find all equilibria.
c) Ascertain the stability of each equilibrium point.
d) Use any appropriate numerical method such as Mathematica’s NDSolve, MATLAB, or a straightforward Eulermethod to integrate the governing differential equation from the initial state to the equilibrium state.
e) Repeat this usingchemkin to generate the reaction rates anddlsode to integrate the differential equations.
f) Plot the Gibbs free energy,G = nH2 gH2 + nH gH , as a function of time.
Formulation of the reaction kinetics:
dρH2
dt= v1aT β e
−ERT
(
N
∏k=1
ρv′kk
)(
1− 1kc
N
∏k=1
ρvkk
)
(323)
a = 2.23·109 m3 ·K− 12
kmol · s , E = 387,623.6 kJ/kmol (324)
(ρH2)t=0 =P0
RT=
1500kPa
8.314 kJkmol·K ·5200K
= 0.034696kmol/m3 (325)
V =(nH2)t=0
(
¯rhoH2
)
t=0
= 28.82m3 (326)
−dnH2 =12
nH →−(
nH2 − (nH2)t=0
)
=12
(
nH2 − (nH2)t=0
)
(327)
nH = 2(1 kmol −nH2) , ¯rhoH = 2
(
1 kmol28.82m3 − ¯rhoH2
)
(328)
dρH2
dt= vH2aT β e
−ERT (ρH2)
v′H2 (ρH)v′H
(
1− 1kc
(ρH2)vH2 (ρH)vH
)
(329)
v′H2= 2, v′H = 0, vH2 = −1, vH = 2 (330)
dρH2
dt= −aT β e
−ERT (ρH2)
2(
1− 1kc
ρ2H
ρH2
)
(331)
41
k(T ) = aT β e−ERT = −2.0532·107 (332)
kc =
(
P0
RT
)∑Ni=1 vi
e(−∆GRT ), ∆G = 179564.2 kJ/kmol ⇒ kc = 0.14723 (333)
dρH2
dt= −2.0532·107· ρ2
H2
(
1− 10.14723
· [2(0.034696− ρH2)]2
ρH2
)
(334)
Find all equilibria:
Equilibria are located at the points wheredρH2
dt = 0.
dρH2
dt= 0 when ρH2 = {0,0.0129,0.0933} (335)
Ascertain stability:
For stability,d2 ρ
H2dt2
must be negative. Therefore,
• ρ = 0 is unstable
• ρ = 0.0129 is stable
• ρ = 0.0933 is unstable and non-physical
42